Magnetic Effects & Magnetism

JEE Main level, single correct, medium

72 of 276

Question 72 of 276

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05m0.05m and 0.2m0.2m from the centre are in the ratio 8:18:1. The radius of coil is _______ .

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Answer: (C) 0.1m0.1m

We know, the magnetic field on the axis of a current carrying circular ring is given by
B=μ04π2NIA(R2+x2)3/2B=\dfrac{{\mu }_{0}}{4\pi }\dfrac{2NIA}{{\left({R}^{2}+{x}^{2}\right)}^{3/2}}
∴B1B2=81=[R2+(0.2)2R2+(0.05)2]3/2∴\dfrac{{B}_{1}}{{B}_{2}}=\dfrac{8}{1}={\left[\dfrac{{R}^{2}+{\left(0.2\right)}^{2}}{{R}^{2}+{\left(0.05\right)}^{2}}\right]}^{3/2}
4[R2+(0.05)2]=[R2+(0.2)2]4\left[{R}^{2}+{\left(0.05\right)}^{2}\right]=\left[{R}^{2}+{\left(0.2\right)}^{2}\right]
4R2−R2=(0.2)2−4×(0.05)24{R}^{2}-{R}^{2}={\left(0.2\right)}^{2}-4\times {\left(0.05\right)}^{2}
4R2−R2=(0.2)2−(0.1)24{R}^{2}-{R}^{2}={\left(0.2\right)}^{2}-{\left(0.1\right)}^{2}
3R2=0.3×0.13{R}^{2}=0.3\times 0.1
R2=(0.1)2⇒R=0.1{R}^{2}={\left(0.1\right)}^{2}\Rightarrow R=0.1

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